<!DOCTYPE html>
<html lang="en-US">
<!--********************************************-->
<!--*       Generated from PreTeXt source      *-->
<!--*                                          *-->
<!--*         https://pretextbook.org          *-->
<!--*                                          *-->
<!--********************************************-->
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<meta name="robots" content="noindex, nofollow">
</head>
<body class="ignore-math">
<h3 class="heading"><span class="type">Paragraph</span></h3>
<ul class="disc">
<li>
<p><dfn class="terminology">Convergence</dfn>  A power series <span class="process-math">\(\displaystyle\sum_{n=0}^{\infty} a_n(x - x_0)^n\)</span> <dfn class="terminology">converges</dfn> at a point <span class="process-math">\(x\)</span> if</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\lim_{m\to\infty}\sum_{n=0}^m a_n(x - x_0)^n
\end{equation*}
</div>
<p class="continuation">exists for that <span class="process-math">\(x\text{.}\)</span>The series certainly converges for <span class="process-math">\(x = x_0\text{;}\)</span> it may converge for all <span class="process-math">\(x\text{,}\)</span> or it may converge for some <span class="process-math">\(x\)</span> and not for others.</p>
</li>
<li>
<p><dfn class="terminology">Absolute Convergence</dfn>  A power series <span class="process-math">\(\displaystyle\sum_{n=0}^{\infty} a_n(x - x_0)^n\)</span> <dfn class="terminology">converges absolutely</dfn> at a point <span class="process-math">\(x\)</span> if</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\sum_{n=0}^{\infty} |a_n(x - x_0)^n|=\sum_{n=0}^{\infty}|a_n||x-x_0|^n
\end{equation*}
</div>
<p class="continuation">converges.<em class="emphasis">If the series converges absolutely, then the series also converges</em>; however, the converse is not necessarily true.</p>
</li>
<li>
<p><dfn class="terminology">Ratio Test</dfn>   Convergence of a power series can often be determined by the ratio test. If <span class="process-math">\(a_n\neq 0\text{,}\)</span> and if for a fixed value of <span class="process-math">\(x\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\textcolor{black}{\lim_{n\to\infty} \left|\frac{a_{n+1}(x - x_0)^{n+1}}{a_n(x - x_0)^n}\right|}=|x-x_0|\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|\textcolor{black}{=L|x-x_0|.}
\end{equation*}
</div>
<p class="continuation">The power series converges absolutely at that value of <span class="process-math">\(x\)</span> if <span class="process-math">\(|x-x_0|&lt;1/L\text{,}\)</span> and diverges if <span class="process-math">\(|x-x_0|&gt;1/L\text{.}\)</span> If <span class="process-math">\(|x-x_0|=1/L\text{,}\)</span> the test is inconclusive.</p>
</li>
<li>
<p>The <dfn class="terminology">radius of convergence</dfn> (about <span class="process-math">\(x_0\)</span>): a nonnegative <span class="process-math">\(\textcolor{blue}{\rho}\text{,}\)</span> such that</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\sum_{n=0}^{\infty} a_n(x-x_0)^n ~~\left\{\begin{array}{ll} \text{\textcolor{blue}{converges absolutely}} &amp; \text{for $\textcolor{magenta}{|x-x_0|&lt;}\textcolor{blue}{\rho}$},\\ \text{diverges} &amp;\text{for $|x-x_0|&gt;\textcolor{blue}{\rho}$.}\end{array}\right.
\end{equation*}
</div>
<p>For a series that converges only at <span class="process-math">\(x_0\text{,}\)</span> <span class="process-math">\(\rho=0\text{;}\)</span>for a series that converges for all <span class="process-math">\(x\text{,}\)</span> <span class="process-math">\(\rho\)</span> is infinite.</p>
<p>If <span class="process-math">\(\rho&gt;0\text{,}\)</span> then the interval <span class="process-math">\(|x-x_0|&lt;\rho\)</span> is called <dfn class="terminology">the interval of convergence</dfn>.</p>
</li>
<li>
<p><dfn class="terminology">A Power Series Defines a Function</dfn>  </p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
f(x)=\displaystyle\sum_{n=0}^{\infty} a_n(x-x_0)^n
\end{equation*}
</div>
<p class="continuation">whose domain is the interval of convergence of the series.</p>
<ul class="circle">
<li>
<p>Within the interval of convergence, <span class="process-math">\(f'(x)\)</span> and <span class="process-math">\(\int f(x)\textrm{d}x\)</span> can be found by term-wise differentiation and integration. For example,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
f'(x)=\sum_{n=0}^{\infty} na_n(x-x_0)^{n-1}=\sum_{\textcolor{blue}{n=1}}^{\infty} na_n(x-x_0)^{n-1},
\end{equation*}
</div>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
f''(x)=\sum_{n=0}^{\infty} n(n-1)a_n(x-x_0)^{n-2}=\sum_{\textcolor{blue}{n=2}}^{\infty} n(n-1)a_n(x-x_0)^{n-2}
\end{equation*}
</div>
</li>
<li><p>The differentiated (integrated) series has <em class="emphasis">the same radius of convergence</em> as the original series.</p></li>
<li><p>However, <em class="emphasis">convergence at an endpoint</em> may be either lost by differentiation or gained through integration.</p></li>
</ul>
<p>These results are important and will be used shortly.</p>
</li>
<li>
<p><dfn class="terminology">Useful Expansions</dfn>   Some power series expansions about 0 that you should be familiar with.Except for the last one, the expansions are valid for all <span class="process-math">\(z\)</span> (i.e. the radius of convergence is <span class="process-math">\(\rho=\infty\)</span>).</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}
\end{equation*}
</div>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\cos(z)=\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k}}{(2k)!},\quad \sin(z)=\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)!}
\end{equation*}
</div>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\textcolor{red}{\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n}~~~~(|z|&lt;1)
\end{equation*}
</div>
</li>
<li>
<p><dfn class="terminology">Identity Property</dfn>  </p>
<ul class="circle">
<li>
<p>If <span class="process-math">\(\displaystyle\sum_{n=0}^{\infty} c_n(x-x_0)^n=0\)</span> for all <span class="process-math">\(x\)</span> in the interval of convergence, then</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
c_n=0\quad \text{ for all } n.
\end{equation*}
</div>
</li>
<li>
<p>If <span class="process-math">\(\displaystyle\sum_{n=0}^{\infty} a_n(x-x_0)^n=\sum_{n=0}^{\infty} b_n(x-x_0)^n\)</span> for all <span class="process-math">\(x\)</span> in the interval of convergence, then</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\textcolor{magenta}{a_n=b_n\quad \text{ for all } n}.
\end{equation*}
</div>
<p>Namely, if two power series are <dfn class="terminology">equal</dfn>, then they must have the same coefficients.</p>
</li>
</ul>
</li>
<li>
<p><dfn class="terminology">Arithmetic of Power Series</dfn>   </p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
f(x)=\displaystyle\sum_{n=0}^{\infty} a_n(x-x_0)^n,\qquad g(x)=\displaystyle\sum_{n=0}^{\infty} b_n(x-x_0)^n
\end{equation*}
</div>
<p class="continuation">whose domain is the interval of convergence of the series.</p>
<ul class="circle">
<li><div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
f(x)\pm g(x)=\sum_{n=0}^{\infty} (a_n\pm b_n)~(x-x_0)^n=\sum_{n=0}^{\infty} c_n(x-x_0)^n.
\end{equation*}
</div></li>
<li>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
f(x)g(x)=\left[\sum_{n=0}^{\infty} a_n(x-x_0)^n\right]\left[\sum_{n=0}^{\infty} b_n(x-x_0)^n\right]=\sum_{n=0}^{\infty} d_n(x-x_0)^n,
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(d_n=a_0b_n+a_1b_{n-1}+\cdots+a_nb_0\text{.}\)</span></p>
</li>
<li><p>If <span class="process-math">\(g(x_0)\neq0\text{,}\)</span> the series for <span class="process-math">\(f(x)/g(x)\)</span> can also be found accordingly.</p></li>
</ul>
</li>
<li>
<p><dfn class="terminology">Analytic at a Point</dfn>   A function <span class="process-math">\(f\)</span> is <dfn class="terminology">analytic at a point <span class="process-math">\(x_0\)</span></dfn> if it can be represented by a power series in <span class="process-math">\(x-x_0\)</span> with a positive or infinite radius of convergence (i.e. <span class="process-math">\(\rho&gt;0\)</span>).In calculus, functions like <span class="process-math">\(e^x\text{,}\)</span> <span class="process-math">\(\cos(x)\text{,}\)</span> <span class="process-math">\(\sin(x)\)</span> can be represented by Taylor series with <span class="process-math">\(|x|&lt;\infty\text{.}\)</span></p>
<p>Taylor series centered at 0 are called <dfn class="terminology">Maclaurin series</dfn>. Show that <span class="process-math">\(e^x\text{,}\)</span> <span class="process-math">\(\cos(x)\text{,}\)</span> <span class="process-math">\(\sin(x)\text{,}\)</span> are analytic at <span class="process-math">\(x=0\text{.}\)</span></p>
</li>
<li><p><dfn class="terminology">Shifting the Summation Index</dfn></p></li>
</ul>
<span class="incontext"><a href="sec5_1.html#p-176" class="internal">in-context</a></span>
</body>
</html>
